A box slides down a slope described by: $$y = 0.05x^2$$ where $x$ is the x coordinate of the slope and y is the y co-ordinate (both in meters). Find the $y$ component of acceleration at $0.4 m$ if the $x$ component is $-1.0m/s^2$
My work:
Let $V_y$, $V_x$, $A_y$, and $A_x$ be the vertical and horizontal components of velocity and acceleration respectfully. Let $t$ be time.
$$A_y = \frac{d(V_y)}{dt}$$ $$A_x = \frac{d(V_x)}{dt}$$
$$\frac{A_y}{A_x} = \frac{\frac{d(V_y)}{dt}}{\frac{d(V_x)}{dt}}$$ $$A_y = \frac{\frac{d(V_y)}{dt}}{\frac{d(V_x)}{dt}} A(x)$$ $$A_y = \frac{d(V_y)}{d(V_x)} A(x)$$
$$y = 0.05x^2$$ $$\implies V_y = 0.1V_x$$ $$\implies \frac{d(V_y)}{d(V_x)} = 0.1$$ $$\implies A_y = 0.1A_x$$ $$\implies Ay = -0.1m/s^2$$
which is incorrect. What am I doing wrong? I do not want a solution, I just need someone to point out my mistake. Thanks!
You go from $y=0.05x^2$ to $V_y=0.1V_x$. However, the time derivative of $x^2$ is $2x\frac{dx}{dt}=2xV_x$, so one $x$ is missing from your solution.
These errors would be a lot easier to see if you used units in the variables and constants: define $C = 0.05\text{ m}^{-1}$. Now if you go from $y=C x^2$ to $V_y = 2C V_x$, you immediately notice there is a problem with units in the other equation (meters/second = (1/meters) * (meters/second)).